**1. More examples of type and cotype **

Let be a Banach space. The possible types are in or , . The possible cotypes are in or , .

Cotype does not pass to quotients. Indeed, has cotype , and every separable Banach space is a quotient of . For instance, has trivial cotype.

The type of is .

** 1.1. Complex interpolation **

Type is equivalent to the map , being bounded. By interpolation, if has type and has type , the interpolation space , , has type such that

Cotype does not work so nicely. It is expressible in terms of a map , where

If is complemented, i.e. is the image of a projector independent on (this is the notion of *-convexity*), interpolation works for cotype.

** 1.2. Super **

Clearly, type and cotype are super properties. A stronger property holds.

Definition 1Say iscrudely finitely representable inif there exists such that every finite dimensional subspace of embeds with distorsion in .

The same proof as for ordinary finite representability gives

Proposition 2is crudely finitely representable in iff is isomorphic to a subspace of an ultrapower .

** 1.3. Type and finite representability **

Theorem 3 (Pisier, Maurey-Pisier 1973)Let be a Banach space. Then has trivial type iff is finitely representable in . has trivial cotype iff is finitely representable in .

Lemma 4Let be independent -valued random variables with a symmetric distribution. Let be independent Bernoulli variables. Then and have the same distribution.

*Proof:* of Theorem 3. Let denote the best constant in the inequality

Then . Let us show that is submultiplicative, i.e. . Indeed, group variables in packages of .

Assume that for all . So for every , there exist points is close to . Go to an ultrapower to find vectors such that . The equality case in the chain of inequalities leading to shows that all norms and for all choices of signs, . Fix signs . If is a unit functional, , thus . In other words, these functionals give an isometry of span onto *.*

Otherwise, there exists such that . Then there exists , , such that . Let . Given a family , split it according to norms,

Then

Summing over gives the wanted estimate.

** 1.4. Ribe’s Theorem **

It is our first nonlinear result.

Theorem 5 (Ribe 1976)If is a uniform homeomorphism, then is crudely finitely representable in and is crudely finitely representable in .

*Proof:* Upgrade the uniform homeomorphism to a bi-Lipschitz map. Use differentiability on finite dimensional subspaces to improve bi-Lipschitz maps to isomorphisms.

This implies that and are not uniformly homeomorphic, an earlier result due to Enflo. Enflo’s proof relies on a metric notion of type, on which Naor and Schechtman, Mendel and Naor have elaborated since.

Definition 6Let be a metric space. Say hasEnflo typeif there is a constant such that for all ,

where changes the sign of the -th coordinate.

Remark 1For a Banach space, Enflo type implies usual type .

** 1.5. Proof of Ribe’s Theorem **

Let be a uniform homeomorphism. Then is large-scale Lipschitz: provided . Go to an ultra product to get rid the the large-scale restriction. This means setting . This is bi-Lischitz: . On every finite dimensional subspace , has points of differentiability. The differential is a linear isomorphism of onto a linear subspace of . So is coarsely finitely representable in , and thus in .

**2. -convexity **

This goes back to Beck and Giesy. They were lookibg for a law of large numbers for Banach space valued random variables.

Lemma 7Let have type . Let be independent centered -valued random variables. Then

*Proof:* Reduce to the case of symmetric distributions by replacing by where is an independent copy of . This produces the factor . Then apply Jensen’s inequality

It implies that

which tends to , this is a law of large numbers.

** 2.1. Application: approximation of convex hulls **

Corollary 8Let be a Banach space with type . Let . Let . There exist such that

*Proof:* Apply the law of large numbers to independent copies of the following random variable . By assumption, there exist , such that . So let with probability .

**3. Factorization **

The following was one of the first results on isomorphism types.

Theorem 9 (Kwapién)If a Banach space has simultaneously type and cotype , then is is isomorphic to a Hilbert space.

Up to now we have uses Bernoulli averages. Why not use Gaussian averages ? Grouping Bernoulli variables (divided by ) in packages of produces variables which are approximately standard Gaussian (central limit theorem). So we can use Gaussian averages instead.

Lemma 10Let , be -valued centered Gaussian vectors. Assume that for all ,

Then for every convex function ,

*Proof:* Pick a common orthogonal basis for the quadratic forms and . Use such a basis to find independant from such that both have the same distribution as .

Let be a finite dimensional normed space. Say a homogeneous function is a *cotype * function if for all ,

where are independent standard Gaussians.

Lemma 11There exists a measure on the unit sphere of such that for all ,

*Proof:* Set ,

These are disjoint convex sets

**I have to run to catch the 4:30 train to Orsay… **